HDU 1087 （DP---最大递增子序列和）

最新推荐文章于 2020-03-19 16:42:15 发布

Twillz

最新推荐文章于 2020-03-19 16:42:15 发布

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分类专栏：动态规划

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本文介绍了一款名为“SuperJumping”的游戏及其背后的算法原理。玩家需在棋盘上跳跃，从起点到终点，每次跳跃的目标必须大于当前点。通过动态规划求解最大得分路径，实现最优解。

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Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39083 Accepted Submission(s): 17958

Problem Description
Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now.

The game can be played by two or more than two players. It consists of a chessboard（棋盘）and some chessmen（棋子）, and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path.
Your task is to output the maximum value according to the given chessmen list.

Input
Input contains multiple test cases. Each test case is described in a line as follow:
N value_1 value_2 …value_N
It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.
A test case starting with 0 terminates the input and this test case is not to be processed.

Output
For each case, print the maximum according to rules, and one line one case.

Sample Input
3 1 3 2
4 1 2 3 4
4 3 3 2 1
0

Sample Output
4
10
3

Author
lcy

Recommend

做法：感觉和HDU 1069差不多
dp[i]表示以第i个结尾的子序列的最大和，
dp[i]= max(dp[j]+a[i]) (i>j)
这也很好地解释了hdu 1069为什么要根据面积大小排序，这一题只能从j跳到I也就是只能由j状态转移到i状态，hdu 1069那题也是，大概这就叫做消除后效性（？）

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#define maxn 100010
using namespace std;
typedef long long ll;
ll dp[maxn];
ll s[maxn];
int main()
{
    int  n;
    while(scanf("%d",&n)==1&&n)
    {
        for(int i=0;i<n;i++)scanf("%I64d",&s[i]);
        ll ans=-1;
        for(int i=0;i<n;i++)
        {
            dp[i]=s[i];
            for(int j=0;j<i;j++)
            {
                if(s[i]>s[j]&&dp[i]<dp[j]+s[i])
                {
                    dp[i]=dp[j]+s[i];
                }
            }
            ans=max(ans,dp[i]);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}